14 General Recursion Formulaa1 = a0a2 = a0/2a3 = a0/6Power SeriesExpansion for exGeneral Recursion FormulaCoefficients of the two series may be equatedonly if the summation limits are the same.Therefore, set m = n+1  n=m-1and m=1 corresponds to n = 0or

17 Solution of the HO Schrödinger EquationRearrangement of the EquationDefine

18 Should we try a Taylor series solution?No!!It would be difficult to satisfy the Boundary Conditions.Instead, it’s better to use:That is, we’ll assume that  is the product of a Taylor Series, H(x),and an asymptotic solution, asympt, valid for large |x|

19 The Asymptotic SolutionWe can show that theasymptotic solution isFor large |x|Plug inNote that asympt satisfied the BC’s;i.e. asympt  0 as x 

20 A Recursion Relation for the General SolutionAssume the general solution is of the formIf one: (a) computes d2/dx2(b) Plugs d2/dx2 and  into the Schrödinger Equation(c) Equates the coefficients of xnThen the following recursion formula is obtained:

21 From the recursion formula:Because an+2 (rather than an+1) is related to an, it is best to considerthe Taylor Series in the solution to be the sum of an even and odd seriesEven SeriesOdd SeriesFrom the recursion formula:Even CoefficientsOdd Coefficientsa0 and a1 are two arbitrary constantsSo far: (a) The Boundary Conditions have not been satisfied(b) There is no quantization of energy

22 Satisfying the Boundary Conditions: Quantization of EnergySo far, there are no restrictions on  and, therefore, none on ELet’s look at the solution so far:Even SeriesOdd SeriesDoes this solution satisfy the BC’s: i.e.   0 as x  ?It can be shown that:unless nTherefore, the required BC’s will be satisfied if, and only if, boththe even and odd Taylor series terminate at a finite power.

23 We cannot let either Taylor series reach x.Even SeriesOdd SeriesWe cannot let either Taylor series reach x.We can achieve this for the even or odd series by settinga0=0 (even series) or a1=0 (odd series).We can’t set both a0=0 and a1=0, in which case (x) = 0 for all x.So, how can we force the second series to terminate at a finitevalue of n ?By requiring that an+2 = 0•an for some value of n.

24 If an+2 = 0•an for some value of n, then the required BC willbe satisfied.Therefore, it is necessary for some value of the integer, n, that:This criterion puts restrictions on the allowed values of  and,therefore, the allowed values of the energy, E.

28 Quantized Energies Zero Point EnergyOnly certain energy levels are allowedand the separation between levels is:EnergyThe classical HO permits any value of E.Zero Point EnergyThe minimum allowed value for theenergy is:The classical HO can have E=0.Note: All “bound” particles have a minimum ZPE.This is a consequence of the Uncertainty Principle.

29 The Correspondence PrincipleThe results of Quantum Mechanics must not contradict those ofclassical mechanics when applied to macroscopic systems.The fundamental vibrational frequency of the H2molecule is 4200 cm-1. Calculate the energy levelspacing and the ZPE, in J and in kJ/mol.ħ = 1.05x10-34 J•sc = 3.00x108 m/s= 3.00x1010 cm/sNA = 6.02x1023 mol-1Spacing:ZPE:

47 The Force Constant (k) The Interpretation of k V(R)i.e. k is the “curvature” of the plot, and represents the “rapidity”with which the slope turns from negative to positive.Another way of saying this is the k represents the “steepness”of the potential function at x=0 (R=Re).

48 Correlation Between k and Bond StrengthV(R)ReV(R)DoDoSmall kSmall DoLarge kLarge DoDo is the Dissociation Energy of the molecule, andrepresents the chemical bond strength.There is often a correlation between k and Do.

51 Vibrational SpectroscopyEnergy Levels and TransitionsSelection Rulen = 1(+1 for absorption,-1 for emission)IR Spectra: For a vibration to be IR active, the dipole momentmust change during the course of the vibration.Raman Spectra: For a vibration to be Raman active, the polarizabilitymust change during the course of the vibration.

56 The n=1  n=2 transition is called a hot band, because itsintensity increases at higher temperatureDependence of hot band intensity on frequency and temperatureMolecule  T R~79Br19F cm oC79Br19F79Br19FH35Cl x10-7H35ClH35Cl

58 Vibrational AnharmonicityReV(R)Harmonic OscillatorApproximation:The effect of including vibrational anharmonicity in treating the vibrationsof diatomic molecules is to lower the energy levels and decrease thetransition frequencies between successive levels.

59 A treatment of the vibrations of diatomic molecules which includes V(R)A treatment of the vibrations of diatomic molecules which includesvibrational anharmonicity [includes higher order terms in V(x)]leads to an improved expression for the energy: is the harmonic frequency and xe is the anharmonicity constant.~Measurement of the fundamental frequency (01) and firstovertone (02) [or the "hot band" (1 2)] permits determinationof  and xe.~See HW Problem

68 Vibrations of Polyatomic MoleculesI’ll just outline the results.If one has N atoms, then there are 3N coordinates. For the i’th. atom,the coordinates are xi, yi, zi. Sometimes this is shortened to: xi. = x,y or z.The Hamiltonian for the N atoms (with 3N coordinates), assumingthat the potential energy, V, varies quadratically with the change incoordinate is:

69 After some rather messy algebra, one can transform the Cartesiancoordinates to a set of “mass weighted” Normal Coordinates.There are 3N-6 Normal Coordinates.Each normal coordinate corresponds to the set of vectors showingthe relative displacements of the various atoms during a given vibration.For example, for the 3 vibrations of water, the Normal Coordinatescorrespond to the 3 sets of vectors you’ve seen in other courses.HOHOHO

70 In the normal coordinate system, the Hamiltonian can be written as:Note that the Hamiltonian is of the form:Therefore, one can assume:and simplify the Schrödinger equation by separation of variables.

71 One gets 3N-6 equations of the form:The solutions to these 3N-6 equations are the familiar HarmonicOscillator Wavefunctions and Energies.As noted above, the total wavefunction is given by:The total vibrational energy is:Or, equivalently:

72 In spectroscopy, we commonly refer to the “energy in wavenumbers”,which is actually E/hc:The water molecule has three normal modes, with fundamentalfrequencies: 1 = 3833 cm-1, 2 = 1649 cm-1, 3 = 3943 cm-1.~What is the energy, in cm-1, of the (112) state (i.e. n1=1, n2=1, n3=2)?

73 The water molecule has three normal modes, with fundamentalfrequencies: 1 = 3833 cm-1, 2 = 1649 cm-1, 3 = 3943 cm-1.~What is the energy difference, in cm-1, between (112) and (100),i.e. E(112)/hc – E(100)/hc ?This corresponds to the frequency of the combination band in whichthe molecule’s vibrations are excited from n2=0n2=1 andn3=0n3=2.